By Connel E.H.

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However, it is easy to do in case H is a normal subgroup, which is described below. Theorem 1) 2) 3) 4) If H is a subgroup of G, then the following are equivalent. , if a ∈ G, ∃ b ∈ G with Ha = bH. Proof 1) ⇒ 2) is obvious. Suppose 2) is true and show 3). We have (aHa−1 )a ⊂ Ha so aH ⊂ Ha. Also a(a−1 Ha) ⊂ aH so Ha ⊂ aH. Thus aH = Ha. 3) ⇒ 4) is obvious. Suppose 4) is true and show 3). Ha = bH contains a, so bH = aH because a coset is an equivalence class. Finally, suppose 3) is true and show 1).

Show that in Z12 the equation x2 = 1 has four ¯ solutions. Finally show that if R is a domain, x2 = 1 can have at most two solutions ¯ in R. Subrings Suppose S is a subset of a ring R. The statement that S is a subring of R means that S is a subgroup of the group R, 1 ∈ S , and (a, b ∈ S ⇒ a · b ∈ S). ¯ Then clearly S is a ring and has the same multiplicative identity as R. Note that Z is a subring of Q, Q is a subring of R, and R is a subring of C. Subrings do not play a role analogous to subgroups.

5) ¯ is a subgroup of G, ¯ f −1 (H) ¯ is a subgroup of G. Furthermore, if H ¯ is If H ¯ then f −1 (H) ¯ is normal in G. normal in G, 6) The kernel of f is a normal subgroup of G. , if f (g) = g¯ then If g¯ ∈ G, −1 f (¯ g ) = N g where N = ker(f ). In other words, if the equation f (x) = g¯ has a Chapter 2 Groups 29 solution, then the set of all solutions is a coset of N = ker(f ). This is a key fact which is used routinely in topics such as systems of equations and linear differential equations.